Prove that a left semi-simple ring $R$ is both left noetherian and left-artinian.

I am following the proof given in pg 27,A first course in non-commutative rings (T.Y.Lam). Its strategy is to show that R has a finite composition series from which the result will follow.

It starts out by observing :-

$R=\bigoplus_{i \in I} I_{i}$ where each $I_{i}$ is simple and hence a minimal left ideal in $R$. Then it says that $1$ belongs to $R$,hence it is easily seen to be FINITE direct sum. I am not able to figure out this easy part.

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    $\begingroup$ An infinite direct sum of nonzero modules is never finitely generated. $\endgroup$ – egreg Apr 17 '15 at 13:21
  • $\begingroup$ @egreg ,i am sorry i didn't get it ,how is this statement relevant? $\endgroup$ – Prayagdeep Parija Apr 17 '15 at 13:24
  • $\begingroup$ An infinite direct sum of (nonzero) $R$-modules is not finitely generated, and $R_R$ is finitely generated. It only takes an application of modus tollens to conclude that $R$ is not an infinite direct sum of nonzero modules. $\endgroup$ – rschwieb Apr 17 '15 at 15:03
  • $\begingroup$ Shouldn't it be p. 29? $\endgroup$ – AgentSmith Apr 26 '18 at 20:55

An infinite direct sum of nonzero modules is never finitely generated. Since $R$, as a left module over itself, is finitely generated, the result follows.

However, in this case it's even easier: you can write $1=\sum_{\lambda\in\Lambda}x_\lambda$, with $x_\lambda\in I_\lambda$ (I changed slightly the notation, with $\Lambda$ as index set) and all but a finite number of $x_\lambda$ different from $0$. Then, for $r\in R$, $$ r=r1=\sum_{\lambda\in\Lambda}rx_\lambda $$ which shows that $r\in\bigoplus_{\lambda\in\Lambda_0}I_\lambda$, where $\Lambda_0=\{\lambda\in\Lambda:x_\lambda\ne0\}$.

Thus $\bigoplus_{\lambda\in\Lambda}I_\lambda\subseteq\bigoplus_{\lambda\in\Lambda_0}I_\lambda$ and we are done.

  • $\begingroup$ i still don't get,if i am understanding correctly, you are proceeding via contradiction,and the last statement is infact a contradiction to the fact that there can't be proper containment,leaving the case when they are equal.what if they are equal.? $\endgroup$ – Prayagdeep Parija Apr 17 '15 at 13:55
  • $\begingroup$ @Prayagdeep No, this is a direct proof, and there is no proof by contradiction here. By showing the last line, egreg has established that the two direct sums are equal, and it follows that $\Lambda=\Lambda_0$. $\endgroup$ – rschwieb Apr 17 '15 at 15:00

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